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## Homework Statement

if: [tex] (1+\sqrt{1+a})\tan{\alpha} = (1+\sqrt{1-a}) [/tex]

Prove that: [tex] \sin{4 \alpha} = a [/tex]

## Homework Equations

[tex] \cos{2\alpha} = 1-2\sin^2{\alpha} [/tex]

[tex] \tan{\alpha} = \sqrt{\frac{1-\cos{2\alpha}}{1+\cos{2\alpha}}}[/tex]

## The Attempt at a Solution

We probably have to simplify to express [tex] \tan{\alpha} [/tex] in terms of [tex]\sin{4\alpha}[/tex] and hence need to use the equations given above. Trying componendo and dividendo did no good as it got too complicated.

[tex]\sin{\alpha}+\sin{\alpha}\sqrt{1+a} = \cos{\alpha}+\cos{\alpha}\sqrt{1-a}[/tex]

[tex](\sin{\alpha}-\cos{\alpha})^2 = (\cos{\alpha}\sqrt{1-a}-\sin{\alpha}\sqrt{1+a})^2 [/tex]

[tex]1-\sin{2\alpha}=\cos^2{\alpha}(1-a)+\sin^2{\alpha}(1+a)-\sin{2\alpha}\sqrt{1-a^2}[/tex]

Though I could simplify after this, I cannot understand how to get [tex]\sin{4\alpha} [/tex] out of it. What is looking even more difficult is how to remove the [tex] \sqrt{1-a^2} [/tex] term.

I would be grateful if somebody could guide me in solving this.